Thu Feb 25 19:24:00 MST 1999
aquarius% maple
|\^/| Maple V Release 5.1 (WMI Campus Wide License)
._|\| |/|_. Copyright (c) 1981-1998 by Waterloo Maple Inc. All rights
\ MAPLE / reserved. Maple and Maple V are registered trademarks of
<____ ____> Waterloo Maple Inc.
| Type ? for help.
# ----------[ M a p l e ]----------
#interface(echo = 3);
# ---------- Initialization ----------
> readlib(showtime):
> on;
# ---------- Statistics ----------
O1 := with(stats):
time = 0.08, bytes = 166254
# Compute the mean of a list of numbers => 9
O2 := describe[mean]([3, 7, 11, 5, 19]);
9
time = 0.06, bytes = 73142
# Compute the median of a list of numbers => 7
O3 := describe[median]([3, 7, 11, 5, 19]);
7
time = 0.02, bytes = 21594
# Compute the first quartile (25% quantile) of a list of numbers => 2 or 5/2
O4 := xx:= [1, 2, 3, 4, 5, 6, 7, 8]:
time = 0.02, bytes = 10038
O5 := describe[quartile[1]](xx);
2
time = 0.01, bytes = 17982
O6 := describe[quantile[1/4]](xx);
2
time = 0.02, bytes = 13298
O7 := xx:= 'xx':
time = 0.00, bytes = 4274
# Compute the mode (the most frequent item) of a list of numbers => 7
O8 := describe[mode]([3, 7, 11, 7, 3, 5, 7]);
7
time = 0.02, bytes = 19010
# Compute the unbiased sample standard deviation of a list of numbers
# => sqrt(5/2)
O9 := describe[standarddeviation[1]]([1, 2, 3, 4, 5]);
1/2
1/2 10
time = 0.05, bytes = 40118
# Discrete distributions---what is the probability of finding exactly 12
# switches that work from a group of 15 where the probability of a single one
# working is 75%? (Need to use the probability density function [PDF] of the
# binomial distribution.) => 0.22520
O10 := statevalf[pf, binomiald[15, .75]](12);
.2251990652
time = 0.04, bytes = 66494
# Replace `exactly' by `up through' in the above. (Need to use the cumulative
# probability density function [CDF] of the binomial distribution.) => 0.76391
O11 := statevalf[dcdf, binomiald[15, .75]](12);
.7639121888
time = 0.03, bytes = 38246
# Continuous distributions---if a radiation emission can be modeled by a
# normal distribution with a mean of 4.35 mrem and a standard deviation of
# 0.59 mrem, what is the probability of being exposed to anywhere from 4 to 5
# mrem? => .5867
O12 := statevalf[cdf, normald[4.35, 0.59]](5)
O12 := - statevalf[cdf, normald[4.35, 0.59]](4);
.5881859844
time = 0.07, bytes = 68442
# Hypothesis testing---how good of a guess is 5 for the mean of xx?
O13 := xx:= [1, -2, 3, -4, 5, -6, 7, -8, 9, 10]:
time = 0.01, bytes = 9762
# Using Student's T distribution (preferred) => 0.057567
O14 := statevalf[cdf, studentst[nops(xx) - 1]]((describe[mean](xx) - 5)/
O14 := (describe[standarddeviation[1]](xx) / sqrt(nops(xx))));
.0575666011
time = 0.07, bytes = 64010
# Using the normal distribution (as an alternative) => 0.040583
O15 := statevalf[cdf, normald]((describe[mean](xx) - 5)/
O15 := (describe[standarddeviation[1]](xx) / sqrt(nops(xx))));
.04058346175
time = 0.06, bytes = 53010
O16 := xx:= 'xx':
time = 0.00, bytes = 3494
# Chi-square test---what is the expectation that row characteristics are
# independent of column characteristics for a two dimensional array of data?
# => 0.469859 (chi2 = 1153/252)
O17 := x:= matrix([[41, 27, 22], [79, 53, 78]]):
time = 0.01, bytes = 26078
O18 := m:= linalg[rowdim](x):
time = 0.02, bytes = 40434
O19 := n:= linalg[coldim](x):
time = 0.01, bytes = 7126
O20 := rowSum:= [sum(x[i, j], j = 1..n) $ i = 1..m]:
time = 0.02, bytes = 21502
O21 := colSum:= [sum(x[i, j], i = 1..m) $ j = 1..n]:
time = 0.01, bytes = 7342
O22 := matSum:= sum(rowSum[i], i = 1..m):
time = 0.01, bytes = 6202
O23 := e:= matrix(m, n, (i, j) -> rowSum[i]*colSum[j]/matSum):
time = 0.01, bytes = 7810
O24 := chi2:= sum(sum((x[i, j] - e[i, j])^2/e[i, j], i = 1..m), j = 1..n);
bytes used=1000024, alloc=851812, time=0.83
1153
----
252
time = 0.09, bytes = 9594
O25 := 1 - statevalf[cdf, chisquare[m*n - 1]](chi2);
.4698594143
time = 0.07, bytes = 76994
O26 := chi2:= 'chi2':
time = 0.01, bytes = 3830
O27 := colSum:= 'colSum':
time = 0.00, bytes = 3934
O28 := matSum:= 'matSum':
time = 0.00, bytes = 3950
O29 := rowSum:= 'rowSum':
time = 0.01, bytes = 4034
O30 := e:= 'e':
time = 0.00, bytes = 3982
O31 := m:= 'm':
time = 0.01, bytes = 3774
O32 := n:= 'n':
time = 0.00, bytes = 4062
O33 := x:= 'x':
time = 0.00, bytes = 3942
# Linear regression (age as a function of developmental score). See Lambert
# H. Koopmans, _Introduction to Contemporary Statistical Methods_, Second
# Edition, Duxbury Press, 1987, p. 459 => y' = 0.7365 x + 6.964
O34 := [[3.33, 3.25, 3.92, 3.50, 4.33, 4.92, 6.08, 7.42, 8.33, 8.00, 9.25,
O34 := 10.75],
O34 := [8.61, 9.40, 9.86, 9.91, 10.53, 10.61, 10.59, 13.28, 12.76, 13.44, 14.27,
O34 := 14.13]]:
time = 0.04, bytes = 31974
O35 := fit[leastsquare[[x, y], y = a + b*x, {a, b}]](%);
y = 6.964118392 + .7364611289 x
time = 0.39, bytes = 722514
# Multiple linear regression (income as a function of age and years of
# college) => y = -16278.7 + 960.925 x1 + 2975.66 x2
O36 := [[37, 45, 38, 42, 31], [4, 0, 5, 2, 4], [31200, 26800, 35000, 30300, 25400]]:
time = 0.02, bytes = 12786
O37 := fit[leastsquare[[x1, x2, y], y = a + b*x1 + c*x2, {a, b, c}]](%);
bytes used=2000352, alloc=1244956, time=1.67
58782300 3469900 10745100
y = --------- + ------- x1 + -------- x2
3611 3611 3611
time = 0.28, bytes = 107234
O38 := evalf(%);
y = -16278.67627 + 960.9249515 x1 + 2975.657713 x2
time = 0.01, bytes = 4554
# Multiple linear regression using the L1 or Least Absolute Deviations
# technique rather than the Least Squares technique (minimizing the sum of the
# absolute values of the residuals rather than the sum of the squares of the
# residuals). Here, the Stack-loss Data is used (percentage of ammonia lost
# times 10 from the operation of a plant over 21 days as a function of air
# flow to the plant, cooling water inlet temperature and acid concentration).
# See W. N. Venables and B. D. Ripley, _Modern Applied Statistics with
# S-plus_, Springer, 1994, p. 218.
# => y = 0.83188 x1 + 0.57391 x2 - 0.06086 x3 - 39.68984
O39 := [[80, 80, 75, 62, 62, 62, 62, 62, 58, 58, 58, 58, 58, 58, 50, 50, 50, 50, 50,
O39 := 56, 70],
O39 := [27, 27, 25, 24, 22, 23, 24, 24, 23, 18, 18, 17, 18, 19, 18, 18, 19, 19, 20,
O39 := 20, 20],
O39 := [89, 88, 90, 87, 87, 87, 93, 93, 87, 80, 89, 88, 82, 93, 89, 86, 72, 79, 80,
O39 := 82, 91],
O39 := [42, 37, 37, 28, 18, 18, 19, 20, 15, 14, 14, 13, 11, 12, 8, 7, 8, 8, 9,
O39 := 15, 15]]:
time = 0.11, bytes = 76106
O40 := fit[leastsquare[[x1, x2, x3, y], y = a + b*x1 + c*x2 + d*x3, {a, b, c, d}]](%);
47136342623 845013447 191180951 44905797
y = ------------ + ---------- x1 + --------- x2 - --------- x3
1180779736 1180779736 147597467 295194934
time = 0.25, bytes = 216962
O41 := evalf(%);
y = -39.91967442 + .7156402005 x1 + 1.295286124 x2 - .1521225191 x3
time = 0.01, bytes = 4398
# Nonlinear regression (Weight Loss Data from an Obese Patient consisting of
# the time in days and the weight in kilograms of a patient undergoing a
# weight rehabilitation program). Fit this using least squares to weight =
# b0 + b1 2^(- days/th), starting at (b0, b1, th) = (90, 95, 120) [Venables
# and Ripley, p. 225] => weight = 81.37375 + 102.6842 2^(- days/141.9105)
O42 := [[ 0, 4, 7, 7, 11, 18, 24, 30, 32, 43, 46, 60, 64, 70, 71,
O42 := 71, 73, 74, 84, 88, 95, 102, 106, 109, 115, 122, 133, 137, 140, 143,
O42 := 147, 148, 149, 150, 153, 156, 161, 164, 165, 165, 170, 176, 179, 198, 214,
O42 := 218, 221, 225, 233, 238, 241, 246],
O42 := [184.35, 182.51, 180.45, 179.91, 177.91, 175.81, 173.11, 170.06, 169.31,
O42 := 165.10, 163.11, 158.30, 155.80, 154.31, 153.86, 154.20, 152.20, 152.80,
O42 := 150.30, 147.80, 146.10, 145.60, 142.50, 142.30, 139.40, 137.90, 133.70,
O42 := 133.70, 133.30, 131.20, 133.00, 132.20, 130.80, 131.30, 129.00, 127.90,
O42 := 126.90, 127.70, 129.50, 128.40, 125.40, 124.90, 124.90, 118.20, 118.20,
O42 := 115.30, 115.70, 116.00, 115.50, 112.60, 114.00, 112.60]]:
time = 0.08, bytes = 82150
# ---------- Quit ----------
O43 := quit
bytes used=2410140, alloc=1376004, time=2.15
real 2.98
user 2.23
sys 0.63